Electrochemistry 3: Cell potentials and thermodynamicsTable of contents
IntroductionIt has long been known that some metals are more "active" than others in the sense that a more active metal can "displace" a less active one from a solution of its salt. The classic example is the one we have already mentioned on the preceding page: Zn(s) + Cu2+ → Zn2+ + Cu(s) Here zinc is more active because it can displace (precipitate) copper from solution. If you immerse a piece of metallic zinc in a solution of copper sulfate, the surface of the zinc quickly becomes covered with a black coating of finely-divided copper, and the blue color of the hydrated copper(II) ion diminishes. Similar comparisons of other metals made it possible to arrange them in the order of their increasing electron-donating (reducing) power. This sequence became known as the electromotive or activity series of the metals.
The activity series has long been used to predict the direction of oxidation-reduction reactions; see here for a nicely-done table with explanatory material. Consider, for example, the oxidation of Cu by metallic zinc that we have mentioned previously. The fact that zinc is near the top of the activity series means that this metal has a strong tendency to lose electrons. By the same token, the tendency of Zn to accept electrons is relatively small. Copper, on the other hand, is a poorer electron donor, and thus its oxidized form, Cu, is a fairly good electron acceptor. We would therefore expect the reaction Zn(s) + Cu2+ → Zn2+ + Cu(s) to proceed in the direction indicated, rather than in the reverse direction. An old-fashioned way of expressing this is to say that "zinc will displace copper from solution". The above table is of limited practical use because it does not take into account the concentrations of the dissolved species. In order to treat these reactions quantitatively, it is convenient to consider the oxidation and reduction steps separately. Standard half-cell potentialsWhen a net reaction proceeds in an electrochemical cell, oxidation occurs at one electrode (the anode) and reduction takes place at the other electrode (the cathode.) We can think of the cell as consisting of two half-cells joined together by an external circuit through which electrons flow and an internal pathway that allows ions to migrate between them so as to preserve electroneutrality. Reduction potentialsEach half-cell has associated with it an electrode-solution potential difference whose magnitude depends on the nature of the particular electrode reaction and on the concentrations of the dissolved electroactive species. The sign of this potential difference depends on the direction (oxidation or reduction) in which the electrode reaction proceeds. In order express them in a uniform way, we adopt the convention that half-cell potentials are always defined for the reduction direction. Thus the half-cell potential for the Zn/Zn2+ electrode (or couple as it is sometimes called) always refers to the reduction reaction Zn2+ + 2e– → Zn(s) In the cell Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) the zinc appears on the left side, indicating that it is being oxidized, not reduced. For this reason, the potential difference contributed by the left half-cell has the opposite sign to its conventional half-cell potential. More generally, we can define the cell potential or cell EMF as Ecell = ΔV = Eright – Eleft (1) in which "right" and "left" refer to the cell notation convention ("reduction on the right") and not, of course, to the physical orientation of a real cell in the laboratory. If we expand the above expression we see that the cell potential Ecell = VCu – Vsolution + Vsolution – VZn is just the difference between the two half-cell potentials Eright and Eleft. Reference half-cellsThe fact that individual half-cell potentials are not directly measurable does not prevent us from defining and working with them. Although we cannot determine the absolute value of a half-cell potential, we can still measure its value in relation to the potentials of other half cells. In particular, if we adopt a reference half-cell whose potential is arbitrarily defined as zero, and measure the potentials of various other electrode systems against this reference cell, we are in effect measuring the half-cell potentials on a scale that is relative to the potential of the reference cell. The reference cell that has universally been adopted for this purpose is the hydrogen half-cell Pt | ½ H2(g) | H+(aq) || ... in which hydrogen gas is allowed to bubble over a platinum electrode having a specially treated surface which catalyzes the reaction ½ H2(g) → H+ + e– When this electrode is operated under standard conditions of 1 atm H2 pressure, 25°C, and pH = 0, it becomes the standard hydrogen electrode, sometimes abbreviated SHE. In order to measure the relative potential of some other electrode couple M2+/M, we can set up a cell Pt | H2(g) | H+ || M2+ (aq) | M(s) whose net reaction is H2(g) + M2+(aq) → 2H+ + M(s) the potential difference between the platinum and M electrodes will be Ecell = VM – Vsolution + Vsolution – V Pt but since the difference Vsolution – V Pt is by definition zero for the hydrogen half-cell, the cell potential we measure corresponds to Ecell = VM – Vsolution which is just the potential (relative to that of the SHE) of the half-cell whose reaction is M2+ + 2e– → M(s)
Standard [reduction] potentials are commonly denoted by the symbol E°. E° values for hundreds of electrodes have been determined (mostly in the period 1925-45, during which time they were referred to as "oxidation potentials") and are usually tabulated in order of increasing tendency to accept electrons (increasing oxidizing power.)
Given the E° values for two half reactions, you can easily predict the potential difference of the corresponding cell: simply add the reduction potential of the reduction half-cell to the negative of the reduction potential (that is, to the oxidation potential) of the oxidation reaction. Example 1Find the standard potential of the cell Cu(s) | Cu2+ || Cl– | AgCl(s) | Ag(s) and predict the direction of electron flow when the two electrodes are connected. Solution: The net reaction corresponding to this cell will be: Cu(s) + 2AgCl(s) → 2 Ag(s) + 2 Cl–(aq) + Cu2+(aq) Where Cu(s)/Cu2+ is being Oxidized and Ag(s)/Ag+ is being Reduced Ecell = Ereduction + Eoxidation Or Written another way Ecell = Eright - Eleft =(.222 + (– .337)) v = -0.115 v Since Ecell is negative, the reaction will run in the opposite direction. The correct net reaction will be: 2 Ag(s) + 2 Cl–(aq) + Cu2+(aq) → 2AgCl(s) + Cu(s) Where Ag(s)/Ag+ is being Oxidized and Cu(s)/Cu2+ is being Reduced Ecell = Ereduction + Eoxidation Or Written another way Ecell = Eright - Eleft =(.337 + (– .222)) v = +0.115 v Since this potential is positive, the reaction will proceed to the right; electrons will be withdrawn from the silver electrode and flow through the external circuit into the copper electrode. Note carefully that in combining these half-cell potentials, we did not multiply E° for the Cu2+/Cu couple by two. The reason for this will be explained later. Cell potentials and Free (Gibbs) energyFrom the above, it should be apparent that the potential difference between the electrodes of a cell is a measure of the tendency for the cell reaction to take place: the more positive the cell potential, the greater the tendency for the reaction to proceed to the right. But we already know that the standard free energy change expresses the tendency for any kind of process to occur under the conditions of constant temperature and pressure. Thus ΔG° and E° measure the same thing, and are related in a simple way:
... or in more detail (see below for explanations of the units given for voltage)
A few remarks are in order about this very fundamental and important relation:
1 J = 1 watt-sec = 1 (amp-sec) × volts
Thus the volt has the dimensions of joules/coulomb– the energy produced per quantity of charge passing through the cell. Because voltage is the quotient of two extensive quantities, it is itself intensive. When we multiply the anodic and cathodic half-reactions by the stoichiometric factors required to ensure that each involves the same quantity of charge, the free energy change and the number of coulombs both increase by the same factor, leaving the potential (voltage) unchanged. This explains why we do not have to multiply the E°s of the anode and cathode reactions by stoichiometric factors when we are finding the potential of a complete cell. If Eq. 2 is solved for E°, we have
This states explicitly that the cell potential is a measure of the free energy change per mole of electrons transferred, which is a brief re-statement of the principle explained immediately above. To see this more clearly, consider the cell Cu(s) | Cu2+ || Cl– | AgCl(s) | Ag(s) for which we list the standard reduction potentials and ΔG°s of the half-reactions:
Here we multiply the cathodic reaction by two in order to balance the charge. Because the anodic reaction is written as an oxidation, we reverse the sign of its E° and obtain Ecell = Eright – Eleft = –.115 volt for the cell potential. The negative cell potential tells us that this reaction will not proceed spontaneously. When the electrons don't cancel outNote, however, that if we are combining two half reactions to obtain a third half reaction, the E° values are not additive, since this third half-reaction is not accompanied by another half reaction that causes the charges to cancel. Free energies are always additive, so we combine them, and use ΔG° = –nFE° to find the cell potential. Example 2Calculate E° for the electrode Fe3+/Fe(s) from the standard potential of the couples Fe3+/Fe2+ and Fe2+/Fe(s) Solution: Tabulate the values and calculate the ΔG°s as follows:
The free energy for half-reaction (iii) is .109nF, so E°3 = –.109/3 = –.036 v The fall of the electronA table of standard half-cell potentials summarizes a large amount of chemistry, for it expresses the relative powers of various substances to donate and accept electrons by listing reduction half-reactions in order of increasing E° values, and thus of increasing spontaneity. The greater the value of E°, the greater the tendency of the substance on the left to acquire electrons, and thus the stronger this substance is as an oxidizing agent. If you have studied elementary chemical thermodynamics, you will have learned about the role that a quantity called the Gibbs free energy, usually referred to as simply the "free energy", plays in determining the direction of any chemical change. The rule is that all spontaneous change (that is, all reactions that proceed to the "right") is associated with a fall in the free energy, and the greater the degree of that fall (ΔG°), the greater will be the tendency for the reaction to take place.
Since oxidation-reduction processes involve the transfer of an electron from a donor to an acceptor, it makes sense to focus on the electron and to consider that it falls from a higher-free energy environment (the reductant, or "source") to a lower-free energy one (the oxidant, or "sink".) As can be seen from the diagram below, this model makes it far easier to predict what will happen when two or more oxidants and reducants are combined; the electron "falls" as far as it can, filling up oxidizing agents (sinks) from the bottom up, very much in the same way as electrons fill atomic orbitals as we build up larger atoms.
A more detailed table with a more complete explanation can be seen on the "Fall of the electron" tutorial page; it is strongly recommended that you take the time to acquire a thorough understanding of this concept. At this point, it might be worth calling your attention to the similar way of depicting acid-base reactions as representing the "fall of the proton" as shown below and described much more thoroughly here.
An important difference between proton transfer and electron transfer reactions is that the latter can vary greatly in speed, from almost instantaneous to so slow as to be unobservable. Acid-base reactions are among the fastest known. Considerable insight into the chemistry of a single element can be had by comparing the standard electrode potentials (and thus the relative free energies) of the various oxidation states of the element. The most convenient means of doing this is the Latimer diagram. As examples, diagrams for iron and chlorine are shown below. The formulas of the species that represent each oxidation state of the element are written from left to right in order of decreasing oxidation number, and the standard potential for the reduction of each species to the next on the right is written in between the formulas. Potentials for reactions involving hydrogen ions will be pH dependent, so separate diagrams are usually provided for acidic and alkaline solutions (effective hydrogen ion concentrations of 1M and 10–14 M, respectively). The more positive the reduction potential, the greater will be the tendency of the species on the left to be reduced to the one on the right. To see how Latimer diagrams are used, look first at the one for iron in acid solution. The line connecting Fe3+ and Fe2+ represents the reaction Fe3+ + e– → Fe2+ whose positive E° (.440 v) indicates that metallic iron will dissolve in acidic solution to form Fe2+. Because the oxidation of this species to the +3 state has a negative potential (-.771v; moving to the left on the diagram reverses the sign), the +2 state will be the stable oxidation state of iron under these conditions. DisproportionationAn important condition to recognize in a Latimer diagram is when the potential on the left of a species is less positive than that on the right. This indicates that the species can oxidize and reduce itself, a process known as disproportionation. As an example, consider Cl2 in alkaline solution. The potential for its reduction to Cl– is sufficiently positive (+1.35 v) to supply the free energy necessary for the oxidation of one atom of chlorine to hypochlorite. Thus elemental chlorine is thermodynamically unstable with respect to disproportionation in alkaline solution, and the same it true of the oxidation product, ClO– (hypochlorite ion). |

) → 2 LiOH(aq) + H2(g)



Behavior of chlorine in water
You should recall that this is exactly analogous to the expansion of an ideal gas. The full amount of work w = PdV is extracted only under the special condition that the external pressure P opposing expansion is only infinitessimally smaller than the pressure of the gas itself. If the gas is allowed to expand into a vacuum (P = 0), no work will be done.